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Homework Statement
What is the mass of the anhydrous compound in Cu(NO_{3})_{2} * 2.5 H_{2}O
, in other words, the Cu(NO_{3})_{2}, if 4.875g of the hydrated compound is used?
The attempt at a solution
What I did was find the formula mass of Cu(NO_{3})_{2} which turned out to be 187.57g
Cu= 1 * 63.55 = 63.55
N= 2 * 14.01 = 28.02
O = 6 * 16.00 = 96.00
The formula mass of the water is 45.05g
H= 2.5(1.01 * 2) = 5.05g
O= 2.5(16.00 * 1) = 40g
From here I set up a ratio:
4.875/232.62 = x/187.57
x = 3.931g of Cu(NO_{3})_{2}
Is this right?
What is the mass of the anhydrous compound in Cu(NO_{3})_{2} * 2.5 H_{2}O
, in other words, the Cu(NO_{3})_{2}, if 4.875g of the hydrated compound is used?
The attempt at a solution
What I did was find the formula mass of Cu(NO_{3})_{2} which turned out to be 187.57g
Cu= 1 * 63.55 = 63.55
N= 2 * 14.01 = 28.02
O = 6 * 16.00 = 96.00
The formula mass of the water is 45.05g
H= 2.5(1.01 * 2) = 5.05g
O= 2.5(16.00 * 1) = 40g
From here I set up a ratio:
4.875/232.62 = x/187.57
x = 3.931g of Cu(NO_{3})_{2}
Is this right?